If the quadratic equation $(1 + a^2)x^2 + 2abx + (b^2 - c^2) = 0$ has equal and real roots, then prove that:
$$b^2 = c^2(1 + a^2)$$
Generated by claude-sonnet-4-6 · 2026-06-15 10:31 · grounding rag
Model Answer
For equal and real roots, discriminant $D = 0$.
Here, $A = (1+a^2)$, $B = 2ab$, $C = (b^2 - c^2)$.
$$D = B^2 - 4AC = 0$$
$$(2ab)^2 - 4(1+a^2)(b^2 - c^2) = 0$$
$$4a^2b^2 - 4(b^2 - c^2 + a^2b^2 - a^2c^2) = 0$$
$$4a^2b^2 - 4b^2 + 4c^2 - 4a^2b^2 + 4a^2c^2 = 0$$
$$-4b^2 + 4c^2 + 4a^2c^2 = 0$$
$$4c^2(1 + a^2) = 4b^2$$
$$\boxed{b^2 = c^2(1+a^2)}$$
Hence proved.
Source: Chapter 4, Section 4.4 – Nature of Roots
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Explanation
- The key condition for equal real roots is discriminant = 0 ($b^2 - 4ac = 0$).
- Carefully identify $A$, $B$, $C$ from the given equation (don't confuse with the $a, b, c$ in the equation itself — use different letters or be explicit).
- Expand neatly and cancel $4a^2b^2$ terms — that's where students usually lose marks.
- Always write "Hence proved" at the end for proof-type questions.